A stone is tied to a string (length = 1.05 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 14.6 % larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

Question

A stone is tied to a string (length = 1.05 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 14.6 % larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

Answer 1

vertical= 1.14*horizontal

mg+mv^2/r=1.14(mv^2/r)

solve for v. Grab you algebra writing pad.

Answer 2

To determine the speed of the stone, we need to analyze the forces acting on it in each scenario: when the circle is horizontal and when it is vertical.

First, let's consider the horizontal circle. In this case, the tension in the string is equal to the centripetal force required to keep the stone moving in a circular path.

The centripetal force is given by the equation:

F = m * v^2 / r

Where:
F is the tension in the string,
m is the mass of the stone,
v is the speed of the stone, and
r is the radius of the circle.

In the horizontal case, the tension in the string is equal to the weight of the stone since there is no vertical acceleration. So, we can rewrite the equation as:

mg = m * v^2 / r

Simplifying, we get:

g = v^2 / r

Where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now let's consider the vertical circle. In this case, the tension in the string is the sum of the weight of the stone (acting downwards) and the centripetal force required to keep the stone moving in a circular path (acting upwards). The equation can be written as:

T = mg + (m * v^2) / r

Given that the maximum tension in the vertical case is 14.6% larger than the tension in the horizontal case, we can set up the following equation:

T_vertical = T_horizontal + 0.146 * T_horizontal

Simplifying, we get:

T_vertical = T_horizontal * (1 + 0.146)

Substituting the equations for tension in each case, we have:

mg + (m * v^2) / r = g + (v^2) / r * (1 + 0.146)

Simplifying further, we get:

mg + (m * v^2) / r = (g + v^2 / r) * (1 + 0.146)

Now, we can substitute the equation for g = v^2 / r from the horizontal case:

mg + (m * v^2) / r = (v^2 / r + v^2 / r) * (1 + 0.146)

Simplifying, we get:

mg + (m * v^2) / r = (2 * v^2 / r) * (1 + 0.146)

mg + (m * v^2) / r = (2.146 * v^2) / r

Now, we can cancel out the mass and the radius terms:

g + (v^2) / r = 2.146 * (v^2) / r

Next, we can rearrange the equation:

g = (2.146 - 1) * (v^2) / r

Simplifying, we get:

g = 1.146 * (v^2) / r

Finally, we can substitute the value of g and solve for v:

9.8 m/s^2 = 1.146 * (v^2) / 1.05 m

Rearranging the equation, we get:

v^2 = 9.8 m/s^2 * 1.05 m / 1.146

Taking the square root of both sides, we can find the speed of the stone:

v = √(9.8 m/s^2 * 1.05 m / 1.146) ≈ 3.27 m/s

Therefore, the speed of the stone is approximately 3.27 m/s.